|
Wintermute11, jam,
RE: G^2 = g c^2 pi^2 , or also as: g = (G^2/c^2 pi^2)
Thank you for the nod of approval on this rather mysterious equation on gravity. Like any equation, it is only a tool, only a model of what may be happening. In this case, it seems to compute out okay.
As I currently understand it, it merely shows how Newton's G is an expression of the proton to proton gravitation relationship combined with the light speed squared constant and pi squared. What it seems to show is that there is a way to understand how this proton to proton "g" ==> with c^2 and pi^2 ==> goes to "G". The pi squared is also there because, I suspect, Newton's G has a pi function in it. But what does this really mean?
Here, I must confess, I am somewhat stumped and need to work on it further. The implication that electromagnetic energy applied to the proton to proton relationship gives us Newton's Gravity is a great temptation to think of it as an early "unified" theory of energy and gravity, but this is not yet so.
What the above equation means to me is something quite simple, but this may not be acceptable to others, as it should not be, until it can be proven empirically. I see mass as a function of energy, that the more energy there is, the smaller the proton to proton relationship becomes, and thus the weaker the gravity. Vice versa, the lower the total energy, the greater the gravity. This can be most easily illustrated by subtracting this "g" from mass, when mass is expressed as a whole where m =1, in Einstein's famous equation:
E = mc^2 is then rewritten as E = (m-g)c^2.
We know E = 90 petajoules (9x10^16 Joules) in our solar environment, as currently accepted in physics. But what if this number E is lower, say in the Crab Nebula neutron star's environment? Then of necessity, if m =1, and c^2 is constant, then g has to become greater. And if g is greater, then G becomes stronger too. Note this is a "what if" question, so we cannot know the answer at this time because we do not know the total energy output of that neutron star.
However (here I am treading on very thin ice), I do not know that from the "kilograms of mass" can be missing the "kilograms of gravity". What the rewritten equation says, in effect, is that "g" is missing from mass to make it whole. Yet, I am not totally satisfied. Per this rewriting of Einstein's equation, "g" would of necessity become kilograms, if E is expressed in Joules, which is m^2 kg s^-2. What bothers me is that when "g" is expressed in SI base units when applied into the equation that converts it into "G", it becomes something totally different.
To illustrate: for E = (m-g)c^2, the SI units are: (E)m^2 kg s^-2 = ((m) kg - (g) kg) (c^2) m^2 s^-2, so (m) and (g) must be in kilograms. (Note that "m-g" results in a number, (1- 5x10^-39), or slightly less than "one", that then gets multiplied by c^2.) However when this is converted into: G^2 = gc^2 pi^2, it becomes in SI units: (G^2) m^6 kg^-2 s^-4 = (g) (?) (c^2)m^2 s^-2 (pi^2), which means, ignoring the dimensionless "pi squared", g has to be (m^4 kg^-2 s^-2). So here lies my conundrum: How can "kg" become "m^4 kg^-2 s^-2" ? So this is why I had said earlier that I do not understand why this equation works, if it does work. Now you know where I am stuck, but as it is not in my nature to give up, the search goes on. (Please note in my original Axiomatic Equation, I had it work out to E = Joules per seconds, or Watts, but that's another story.)
Perhaps a better understanding of SI base units would help, as per: http://physics.nist.gov/cuu/Units/units.html
Anybody recognize these SI units for the proton to proton "g" as anything? What relationship does "m^4 kg^-2 s^-2" have with "kg"? Do I need a "fudge" factor, or simply forget it? But it's a sickness of my mind, I can't give up, nor forget it! What am I doing wrong? Hmmm... Just had a thought while writing this: "If "m^3 kg^-1 s^-2" is the gravitational force acting on the mass being weighed, then we have simply taken shorthand by converting this into Earth's gravity weight as "kg". So now, if ... if I divide (g) by (kg), and "kg" is really "m^3 kg^-1 s^-2", then the result is "m kg^-1"... but what does "meters per kilogram" mean? Will have to work on it some more.
Thanks for putting up with my "lunatic" equations. Obviously this is still work in progress... or regress. 
With regards to "gravitons", I'm also of the opinion that they are a kind of "shadow" particle of electromagnetic energy, but do not know really if this is true or not, or even if they exist. I think they should exist and travel at v = c, if for no other reason than e.m. waves need to interact somehow with space, which may have a "Casimir" like gravity to it. However, this for me is still in a very speculative stage.
C2
Dark Gravity? See article at NewScientist.com titled:
"Astronomers find first 'dark galaxy'" at:
http://www.newscientist.com/news/news.jsp?id=ns99994272
Cold space, per my theory, should have greater gravity (per mass), and here is some preliminary evidence that it does.
[Edited by Coppernicus2 on 10/23/03 at 02:46]
__________________
I have formally 'resigned' (tactical withdraw) from the Space-Talk boards; mine were many questions, ideas, but no real answers. Thanks. 04/10/04.
Disclaimer: Please note the ideas expressed here by me are cutting edge theory, very speculative in nature, and not physics as it is being currently taught. Caveat lector.
|
Thread 
10/18/03 17:38
